Opposite Sides and Angles of Parallelogram are Equal

Theorem

The opposite sides and angles of a parallelogram are equal to one another, and either of its diameters bisects its area.

Proof

Let $ACDB$ be a parallelogram, and let $BC$ be a diameter.

By definition, $AB \| CD$, and $BC$ intersects both.

So by Parallel Implies Equal Alternate Interior Angles, $\angle ABC = \angle BCD$.

Similarly, by definition, $AC \| BD$, and $BC$ intersects both.

So by Parallel Implies Equal Alternate Interior Angles, $\angle ACB = \angle CBD$.

So $\triangle ABC$ and $\triangle DCB$ have two angles equal, and the side $BC$ in common.

So by Triangle Angle-Side-Angle Equality, $\triangle ABC = \triangle DCB$.

So $AC = BD$ and $AB = CD$.

Also, we have that $\angle BAC = \angle BDC$.

Also, because $\angle ACB = \angle CBD$ and $\angle ABC = \angle BCD$, we have by Common Notion 2 that $\angle ACB + \angle BCD = \angle ABC + \angle CBD$.

So $\angle ACD = \angle ABD$.

So we have shown that opposite sides and angles are equal to each other.

Now note that as $AB = CD$, and $BC$ is common, and $\angle ABC = \angle BCD$, we have that $\triangle ABC = \triangle BCD$ by Triangle Side-Angle-Side Equality.

So $BC$ bisects the parallelogram. Similarly, $AD$ also bisects the parallelogram.

$\blacksquare$

Note

The use of Triangle Side-Angle-Side Equality in this proof seems to be superfluous as the triangles were already shown to be equal using Triangle Angle-Side-Angle Equality. However, Euclid included the step in his proof, so the line is included here.

Note that in at least some translations of The Elements, the Triangle Side-Angle-Side Equality proposition includes the extra conclusion that the two triangles themselves are equal whereas the others do not explicitly state this, but since Triangle Side-Angle-Side equality is used to prove the other congruence theorems, this conclusion would seem to be follow trivially in those cases.

Historical Note

This is Proposition 34 of Book I of Euclid's The Elements.